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Comment on Factorial Notation
Which of the following is a
a)2!+3
b) 4!+12
c)5!+6
d)5!+30
e)8!+12
How do i tackle this problem Brent? Do i have to consider prime numbers?
Nice question!
Nice question!
If A and a multiple of B, then we can write A = kB for some integer k.
For example, we know that 24 is a multiple of 6, because we can write 24 = (4)(6)
4! + 6 = (4)(3)(2)(1) + 6
All MULTIPLES of 4! + 6 are in the form k(4! + 6), where k is an integers
k(4! + 6) = k[(4)(3)(2)(1) + 6]
= (k)(4)(3)(2)(1) + 6k
So, all multiples of 4! + 6 can be written in the form (k)(4)(3)(2)(1) + 6k
At this point, I SCAN the answer choices to see if any of them can be derived from the expression (k)(4)(3)(2)(1) + 6k
Notice that answer choice D, 5! + 30, can be derived if we let k = 5
That is, 5[4! + 6] = 5[(4)(3)(2)(1) + 6]
= (5)(4)(3)(2)(1) + 30
= 5! + 30
Answer: D
Did you work backwards and
Great question.
Great question.
I have edited my response above to address that question.
Cheers.
Brent
Even option E can be written
Be careful. 2(4! + 6) is not
Be careful. 2(4! + 6) is not equivalent to 8! + 12
When we take 2(4! + 6) and EXPAND it, we get:
2(4! + 6) = (2)(4!) + (2)(6)
= (2)(4!) + 12
So the 12 part is correct.
However, (2)(4!) does not equal 8!
(2)(4!) = (2)(4)(3)(2)(1)
8! = (8)(7)(6)(5)(4)(3)(2)(1)
Another way to show that 2(4! + 6) is not equal to 8! + 12 is to EVALUATE each expression.
(2)(4! + 6) = (2)[(4)(3)(2)(1) + 6] = 2[24 + 6] = 2[30] = 60
8! + 12 = (8)(7)(6)(5)(4)(3)(2)(1) + 12 = 40,320 + 12 = 40,332
Cheers,
Brent
Hey Brent,
Came across this question and found that you didn't cover the particular formula for circular arrangements in the module. Can you explain using the FCP?
"How many ways can six friends be arranged around a circular dinner table?"
Thanks
I'm not a big fan of
I'm not a big fan of questions involving people seated at a circular table. Those questions require some assumptions that are un-GMAT-like.
I've seen test-prep companies ask those question types, but I've never seen an official question with the same setup.
Can this be asked on the GMAT
How many zeros does 100! end with?
That would be a reasonable
That would be a reasonable question.
Check out my solution to a VERY SIMILAR question: https://gmatclub.com/forum/if-60-is-written-out-as-an-integer-with-how-m...
The question above involves finding the number of zeros at the end of 60!
See if you can extend that solution to find the number of zeros at the end of 100!
If you need more help, let me know.
Cheers,
Brent
Hi Brent,
Need your help https://gmatclub.com/forum/find-the-number-of-zeroes-in-232903.html
Link: https://gmatclub.com
Link: https://gmatclub.com/forum/find-the-number-of-zeroes-in-232903.html
This is a bad question for 2 reasons:
1) The question should be asking for the number of TRAILING zeros (zeros at the END of the integer). Otherwise, we should also count any other zeros in the number. For example, the number 106,400 has two TRAILING zeros at the end, but there are actually three zeros altogether.
2) The question involves way too many calculations.
KEY CONCEPT: We get ONE trailing zero for every pair of one 2 and one 5 that's hiding in the prime factorization of 1142! × 348! × 17!
So, for example 17! has three 5's in its prime factorization, and 17! has fifteen 2's in its prime factorization
So, in the prime factorization of 17!, there are 3 PAIRS consisting of one 2 and one 5.
As such, 17! has THREE trailing zeros.
This concept can be expanded to 1142! × 348! × 17!
Cheers,
Brent
https://gmatclub.com/forum/if
We are told in algebra when you have the difference of two squares, trick is to factor it out quickly into a^2 - b^2 = (a + b) (a - b)
But that's not the case here.
17! ^ 2 - 16! ^ 2
which should equal 16! (17 + 1) (17 - 1) = 16! (16)(18)
is not the case. Or have I missed something here?
Question link: https:/
Question link: https://gmatclub.com/forum/if-n-n-2-then-162389.html
You've only factored out ONE 16!, when you can actually factor out 16! TWICE.
We have: (17!)² - (16!)² = (17! + 16!)(17! - 16!)
= (16!)(17 + 1)(16!)(17 - 1)
= (16!)²(17 + 1)(17 - 1)
= (16!)²(17² - 1)
= (16!)²(289 - 1)
= (16!)²(288)
= (16!)²(288)
Since we can write: 288 = (12²)(2), we get:
= (¡16!)(12²)(2)
Answer: E
Does that help?
Cheers,
Brent
Aren't we double factoring in
If we have 16! on the outside, doesn't that multiply to all the brackets?
So if we have (16!) on the outside, wouldn't that multiply to BOTH the brackets (17 + 1) and (17 - 1)? In multiplication, isn't it the case that we have to multiply across all the values?
Or have I misunderstood this?
The easiest way to test these
The easiest way to test these kinds of theories is to use actual numbers.
For example, take 8 + 20 and factor it to get: 8 + 20 = 4(2 + 5)
When we evaluate both sides, we get: 28 = 28 (perfect!)
NOTE: the above example involves factoring a SUM. Now let's see what happens when we factor a PRODUCT.
Here's some factoring that's analogous to your factoring:
Take: (8)(20)
Factor 4 from EACH bracket to get: (8)(20) = 4[(2)(5)]
When we evaluate both sides, we get: 160 = 40 (oops!)
So, when it comes to PRODUCTS, we can't factor this way.
Instead, we must factor EACH COMPONENT of the product.
8 = (4)(2)
20 = (4)(5)
So, (8)(20) = (4)(2)(4)(5)
= (4)(4)(2)(5)
= (16)(2)(5)
We get: (8)(20) = (16)(2)(5)
When we evaluate both sides, we get: 160 = 160 (perfect!)
Does that help?
Cheers,
Brent
Are there any rules for
Cheers,
Philipp
When dividing factorials, we
When dividing factorials, we can often cancel out terms.
For example, 7!/5! = (7)(6)(5)(4)(3)(2)(1)/(5)(4)(3)(2)(1)
= (7)(6)
= 42
No nice rule for multiplying factorials.
When adding or subtracting factorials, you might be able to simplify things by factoring.
For example: 6! - 4! = (6)(5)(4)(3)(2)(1) - (4)(3)(2)(1)
= (4)(3)(2)(1)[(6)(5) - 1]
= (4)(3)(2)(1)[30 - 1]
= (4!)[30 - 1]
No nice rule for powers of factorials.
Cheers,
Brent
Hi Brent,
Regarding this question, https://gmatclub.com/forum/if-n-1-n-2-n-440-what-is-the-sum-of-the-digits-of-n-295886.html
How did you go about factoring this quickly?
Question link: https:/
Question link: https://gmatclub.com/forum/if-n-1-n-2-n-440-what-is-the-sum-of-the-digit...
Good question!
I wouldn't exactly say that I factored the quadratic quickly :-)
In fact, the difficult values in the given quadratic make this question a little less GMAT-like.
Given: n² + 4n - 437 = 0
It doesn't take long to realize that there aren't many pairs of values that have a product of 437.
I initially tested whether 437 is divisible by the first five primes (2, 3, 5, 7, and 11), but none of these values work.
So I just kept going (not really any other alternatives).
13...NO GOOD
19...Finally!!!
So, I was able to get: (n + 23)(n - 19) = 0
Cheers,
Brent
Hi Brent,
Do we need to know the circular permutations for the test?
Great question!
Great question!
While a lot of test prep companies have written tons of questions about circular arrangements, I don't believe I've ever seen an OFFICIAL GMAT question test this. That's why I don't include those questions in the Reinforcement Activities boxes.
Hey Brent, in the Solution
Ex. [6 + 3 = (6)(5)(4)(3)(2)(1) + 2 = 3[(6)(5)(4)(2)(1) + 1], which is a multiple of 3. So, [6 + 2 is NOT prime
(sh0ws plus 2 instead of 3)
https://gmatclub.com/forum/for-any-integer-n-greater-than-1-n-denotes-the-product-of-168575.html
Thanks for the heads up!
Thanks for the heads up!
I have edited my response here https://gmatclub.com/forum/for-any-integer-n-greater-than-1-n-denotes-th...
Hi Brent,
https://gmatclub.com/forum/if-n-1-n-2-n-440-what-is-the-sum-of-the-digits-of-n-295886.html
For questions that involve factors that are prime, is there any easy way to determine the factors? Or we always rely on trial and error?
Thanks!
Question link: https:/
Question link: https://gmatclub.com/forum/if-n-1-n-2-n-440-what-is-the-sum-of-the-digit...
We want to factor the following quadratic equation: n² + 4n - 437 = 0
In many cases, we can find the prime factorization of the constant term (the term without any variables), and then use that to help us factor the equation.
However, in this case, that won't help us since the two numbers we're looking for (numbers that multiply to get -437 and add to get 4) are both primes.
In retrospect, this question is beyond the scope of the GMAT. The test makers aren't interested in having students randomly test such large and hard-to-factor numbers. As such, I've removed the question from the Reinforcement Activities